I forgot how to do infinite series how does $$1+e^{-\beta \varepsilon}+e^{-2\beta \varepsilon}+\dots = \frac{1}{1-e^{-\beta \varepsilon}}$$

I don't remember how to do infinite series either tbh

i usually just google them

@Obliv $\frac1{1-x}=1+x+x^2+x^3+\cdots$ and then substitute $x=e^{-\beta\varepsilon}$

Oh thank u, whats the name of this type of series btw?

nvm i got it, it's a geometric series

I hope this is the worst we'll get for partition functions

geometric series that is

for the high temperature limit $k_BT\gg \varepsilon$, how come $\beta \varepsilon \ll 1$ means $e^{\beta \varepsilon}\approx 1+\beta\varepsilon$?

@Obliv For small $|x|$, we get $\ln(1+x)=x$. There are various ways to show that limit. Eg, integrate that geometric series $\frac1{1-x}=1+x+x^2+x^3+\cdots$

Ohh right

It's a very handy limit. Similar to $\lim_{x\to 0} \sin(x) = x$

Also see en.wikipedia.org/wiki/Bernoulli%27s_inequality

thank you. I see it all the time I just forget because I don't know the name of it. I guess just first order taylor approximation?

like $e^x = 1 + \frac{x}{1!}+\frac{x^2}{2!}+...$

Yep

there's also another one I see when we pull out terms like $\sqrt{x^2+a^2}=x\sqrt{1+\frac{a^2}{x^2}}$ where if $x\gg a$ or something then this approximates to something..

perhaps I wrote it wrong, i'll look it up lol

it's $(1+x)^a \approx 1+ax$

i guess for sufficiently small $x$ in that situation

Right. $(1+x/2)^2 = (1 + x + x^2/4)$ So for small $x$ we can ignore the $x^2$ term.

ooh

i'm guessing this only works for exponents $\geq 2$?

Hence $\sqrt{1+x} \approx 1 + x/2$

nvm with higher exponents means more terms to ignore as well..

@Obliv no, this is a general binomial theorem and will work for much more general cases

Veritasium had a video about how Newton changed the game about finding the value of $\pi$, and in it there is a lot of generalisation of the binomial theorem right there.

Also, $(1+x)(1-x)=1-x^2$, so for small $x$, we get $1/(1+x) \approx 1-x$. Which is related to that geometric series.